A-Level · Chemistry · AQA · Mark scheme decoded
AQA A-Level Chemistry: Entropy and Gibbs Free Energy in Chemical Reactions — mark scheme explained
The short answer
In the realm of physical chemistry, understanding the feasibility of chemical reactions is crucial. While enthalpy change (∆H) provides insight into the energy changes during a reaction, it alone is not sufficient to determine whether a reaction will occur spontaneously.
The question
Calculate the entropy change (∆S) for the reaction: H 2 (g) + 1/2 O 2 (g) → H 2 O(g), given the absolute entropy values: S ° (H 2 ) = 130.68 J K -1 mol -1 , S ° (O 2 ) = 205.0 J K -1 mol -1 , and S ° (H 2 O) = 188.84 J K -1 mol -1 .
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for correct substitution of values into formulas, accurate arithmetic, and providing the final answer with appropriate units. For conceptual questions, marks are given for clear and concise explanations that demonstrate a deep understanding of the topic.
What the command words demand
- Calculate
- Perform a numerical calculation using given data and appropriate formulas.
- Determine
- Find or derive a specific value or quantity based on given information.
- Explain
- Provide a detailed description of the concept, including relevant principles and relationships.
- Interpret
- Analyze and explain the significance of thermodynamic data in the context of reaction feasibility.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate approximately 5-7 minutes per question to ensure you have enough time to carefully read the problem, perform calculations, and check your work.
- Calculate the total entropy of the reactants: S reactants ° = 130.68 J K -1 mol -1 + 0.5 × 205.0 J K -1 mol -1 = 233.18 J K -1 mol -11 mark
- Calculate the total entropy of the products: S products ° = 188.84 J K -1 mol -10 marks
- Calculate the entropy change: ∆S = S products ° - S reactants ° = 188.84 J K -1 mol -1 - 233.18 J K -1 mol -1 = -44.34 J K -1 mol -12 marks
Final answer: -44.34 J K -1 mol -1
Work through every step correctly and you earn all 3 marks.
Another worked example
Determine if the reaction H 2 (g) + 1/2 O 2 (g) → H 2 O(g) is feasible at 298 K, given ∆H = -241.8 kJ mol -1 and ∆S = -44.34 J K -1 mol -1 .
- Convert ∆H to joules: ∆H = -241.8 × 10 3 J mol -10 marks
- Calculate T∆S: T∆S = 298 K × (-44.34 J K -1 mol -1 ) = -13,213.32 J mol -11 mark
- Calculate ∆G: ∆G = ∆H - T∆S = -241.8 × 10 3 J mol -1 - (-13,213.32 J mol -1 ) = -228,586.68 J mol -12 marks
Final answer: -228,586.68 J mol -1 (or -228.6 kJ mol -1 )
Work through every step correctly and you earn all 3 marks.
Common mistakes
Using the wrong units for entropy (S) and temperature (T).
Why it happens: Students often forget to convert temperatures to Kelvin or use incorrect units for entropy, leading to incorrect calculations.
Fix: Always ensure that entropy is in joules per kelvin per mole (J K -1 mol -1 ) and temperature is in Kelvin when using the Gibbs free energy equation.
Forgetting to convert enthalpy (∆H) from kJ to J.
Why it happens: The factor of 10 3 can be easily overlooked, leading to incorrect calculations.
Fix: Always convert ∆H from kilojoules (kJ) to joules (J) when substituting into the Gibbs free energy equation.
Confusing the signs of ∆H and ∆S in the Gibbs free energy equation.
Why it happens: Students sometimes mix up the signs of ∆H and ∆S, leading to incorrect values for ∆G.
Fix: Remember that a positive ∆H indicates an endothermic reaction, while a negative ∆H indicates an exothermic reaction. A positive ∆S indicates an increase in disorder, while a negative ∆S indicates a decrease in disorder.
Misinterpreting the feasibility of a reaction based on ∆G.
Why it happens: Students may not fully understand that for a reaction to be feasible, ∆G must be zero or negative (∆G ≤ 0).
Fix: Always check if ∆G is less than or equal to zero to determine the feasibility of a reaction. If ∆G is positive, the reaction is non-spontaneous.
Incorrectly calculating T∆S in the Gibbs free energy equation.
Why it happens: Students may make arithmetic errors or forget to multiply temperature (T) by entropy change (∆S).
Fix: Double-check your calculations for T∆S and ensure that you are multiplying the correct values. Use a calculator if necessary to avoid simple mistakes.
Failing to explain the significance of entropy in determining reaction feasibility.
Why it happens: Students may not fully understand how changes in entropy affect the spontaneity of a reaction, leading to vague or incorrect explanations.
Fix: Practice explaining the role of entropy in reaction feasibility. A positive ∆S indicates an increase in disorder, which is generally favorable for a reaction to proceed. Conversely, a negative ∆S suggests a decrease in disorder, making the reaction less likely to occur spontaneously.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate Entropy Change | Work out the reaction's entropy change from absolute entropy values of reactants and products. | 3 |
| Gibbs Free Energy Feasibility | Calculate ∆G from ∆H and ∆S to decide if the reaction is feasible. | 3 |
| Feasibility Temperature Calculation | Find the temperature at which a reaction becomes feasible using the free energy equation. | 3 |
| Calculate Entropy Change | Use absolute entropy values to work out the entropy change for a reaction. | 3 |
| Total across these question types | 12 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.